Suppose $B \subset \overline{\mathbb{C}}$ is an open set not containing the point at infinity, $v \in C(\overline{B})$ is valued in $[-1,1]$, and Lipschitz on compact subsets of $B$ and $\omega$ is the restriction of a compactly supported smooth function to $\overline{B}$ (in particular it is also Lipschitz on compact subsets of $B$).
Define $v_{\epsilon}(z) = \begin{cases} 1 + \omega(z)&\text{if}\, v(z) \geq \omega(z) + 1 - \epsilon\\ \frac{v(z) - 2 \epsilon \omega(z) - \epsilon}{1 - 2 \epsilon}&\text{if}\, \omega(z) + \epsilon \leq v(z) \leq \omega(z) + 1 - \epsilon\\ \omega&\text{if}\, \omega(z) - \epsilon < v(z) < \omega(z) + \epsilon \\ \frac{v(z) - 2 \epsilon \omega(z) - \epsilon}{1 - 2 \epsilon}&\text{if}\, \omega(z) - 1 + \epsilon \leq v(z) \leq \omega(z) - \epsilon\\ -1 + \omega(z)&\text{if}\, v(z) \leq w(z) - 1 + \epsilon \end{cases}$
Then $v_{\epsilon}$ is continuous on $\overline{B}$ and Lipschitz on compact subsets of $B$ (here $\epsilon > 0$ is supposed to be small).
I want to show that on the boundaries of each of the pieces this function is defined on, $\nabla v_{\epsilon} $ is equal to the gradient of the common function both pieces are equal to almost everywhere, and that in the limit (as $\epsilon \rightarrow 0$) this equals $\nabla(v(z))$.
For instance, on $\{z : v(z) = \omega(z) - 1 + \epsilon \}$, I want to show that $\nabla v_{\epsilon} = \nabla (-1 + \omega(z)) = \nabla(\omega(z)) \rightarrow_{\epsilon \rightarrow 0} \nabla(v(z))$ almost everywhere.
This sounds similar to the answer given in An identity involving the Dirichlet integral , but I am not sure how to apply the solution to this particular case.